Answer :
Solving by factoring in steps shown below
Question 1
- 5x² = 3x + 2
- 5x² - 3x - 2 = 0
- 5x² - 5x + 2x - 2 = 0
- 5x(x - 1) + 2(x - 1) = 0
- (5x + 2)(x - 1) = 0
- 5x + 2 = 0 and x - 1 = 0
- 5x = - 2 and x = 1
- x = - 2/5 and x = 1
Question 2
- x² - 4x + 3 = 0
- x² - x - 3x + 3 = 0
- x(x - 1) - 3(x - 1) = 0
- (x - 3)(x - 1) = 0
- x - 3 = 0 and x - 1 = 0
- x = 3 and x = 1
Answer:
[tex]\textsf{1. \quad $x=1, \quad x=-\dfrac{2}{5}$}[/tex]
[tex]\textsf{2. \quad $x=1, \quad x=3$}[/tex]
Step-by-step explanation:
Solving quadratic equations by factoring
- To factor a quadratic in the form [tex]ax^2+bx+c[/tex] , find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex], and rewrite [tex]b[/tex] as the sum of these two numbers.
- Factor the first two terms and the last two terms separately.
- Factor out the common term.
- Solve for x by applying the zero-product property.
Question 1
Given equation:
[tex]5x^2=3x+2[/tex]
Subtract (3x + 2) from both sides:
[tex]\implies 5x^2-(3x+2)=3x+2-(3x+2)[/tex]
[tex]\implies 5x^2-3x-2=0[/tex]
Find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex], and rewrite [tex]b[/tex] as the sum of these two numbers:
[tex]\implies ac=5 \cdot -2=-10[/tex]
[tex]\implies b=-3[/tex]
Therefore, the two numbers are -5 and 2.
Rewrite [tex]b[/tex] as the sum of the two numbers:
[tex]\implies 5x^2-5x+2x-2=0[/tex]
Factor the first two terms and the last two terms separately:
[tex]\implies 5x(x-1)+2(x-1)=0[/tex]
Factor out the common term (x - 1):
[tex]\implies (5x+2)(x-1)=0[/tex]
Apply the zero-product property:
[tex]\implies 5x+2=0 \implies x=-\dfrac{2}{5}[/tex]
[tex]\implies x-1=0 \implies x=1[/tex]
Therefore, the solutions are:
[tex]\boxed{x=1, \quad x=-\dfrac{2}{5}}[/tex]
Question 2
Given equation:
[tex]\implies x^2-4x+3=0[/tex]
Find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex], and rewrite [tex]b[/tex] as the sum of these two numbers:
[tex]\implies ac=1 \cdot 3=3[/tex]
[tex]\implies b=-4[/tex]
Therefore, the two numbers are -3 and -1.
Rewrite [tex]b[/tex] as the sum of the two numbers:
[tex]\implies x^2-3x-x+3=0[/tex]
Factor the first two terms and the last two terms separately:
[tex]\implies x(x-3)-1(x-3)=0[/tex]
Factor out the common term (x - 3):
[tex]\implies (x-1)(x-3)=0[/tex]
Apply the zero-product property:
[tex]\implies x-1=0 \implies x=1[/tex]
[tex]\implies x-3=0 \implies x=3[/tex]
Therefore, the solutions are:
[tex]\boxed{x=1, \quad x=3}[/tex]